Find a so that lim _(x arrow 1)(x) exist when [ f(x)={ 3 x+5, & x leq 1 2 x+1, & x geqslant 1 . ]

To ensure that the limit of \( f(x) \) as \( x \) approaches 1 exists, we need to make sure that the left-hand limit and the right-hand limit are equal.<br /><br />First, let's find the left-hand limit as \( x \) approaches 1 from the left:<br />\[<br />\lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^-}} (3x + 5) = 3(1) + 5 = 8<br />\]<br /><br />Next, let's find the right-hand limit as \( x \) approaches 1 from the right:<br />\[<br />\lim_{{x \to 1^+}} f(x) = \lim_{{x \to 1^+}} (2x + 1) = 2(1) + 1 = 3<br />\]<br /><br />For the limit to exist, the left-hand limit and the right-hand limit must be equal. Therefore, we need to find a value of \( a \) such that:<br />\[<br />\lim_{{x \to 1}} f(x) = a<br />\]<br /><br />Since the left-hand limit is 8 and the right-hand limit is 3, we need to set \( a \) to be the value that both limits approach. In this case, there is no value of \( a \) that can make the left-hand limit and the right-hand limit equal, as they are already fixed at 8 and 3 respectively.<br /><br />Therefore, the limit of \( f(x) \) as \( x \) approaches 1 does not exist.